Prove that, for n = 1, 2, 3, ... , ∫^2π 0 cos^2n θ dθ = 1 . 3 . 5 ... (2n - 1)/2 . 4 . 6... (2n) 2π

Question

Prove that, for n = 1, 2, 3, ... ,

[latex]\int_0^{2 \pi} \cos ^{2 n} heta d heta=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} 2 \pi[/latex]

Accepted Answer

Let [latex]z=e^{i heta}[/latex]. Then, [latex]d z=i e^{i heta} d heta=i z d heta \quad[/latex] or [latex]\quad d heta=d z / i z \quad[/latex] and [latex]\quad \cos heta=\frac{1}{2}\left(e^{i heta}+e^{-i heta} ight)=\frac{1}{2}(z+1 / z)[/latex]. Hence, if C is the unit circle |z|=1, we have

[latex]\begin{aligned}\int_0^{2 \pi} \cos ^{2 n} heta d heta &=\oint_C\left\{\frac{1}{2}\left(z+\frac{1}{z} ight) ight\}^{2 n} \frac{d z}{i z} \&=\frac{1}{2^{2 n}} \oint_C \frac{1}{z}\left\{z^{2 n}+\left(\begin{array}{c} 2 n \1\end{array} ight)\left(z^{2 n-1} ight)\left(\frac{1}{z} ight)+\cdots+\left(\begin{array}{c}2 n \k\end{array} ight)\left(z^{2 n-k} ight)\left(\frac{1}{z} ight)^k+\cdots+\left(\frac{1}{z} ight)^{2 n} ight\} d z \&=\frac{1}{2^{2 n} i} \oint_C\left\{z^{2 n-1}+\left(\begin{array}{c} 2 n \1\end{array} ight) z^{2 n-3}+\cdots+\left(\begin{array}{c}2 n \k\end{array} ight) z^{2 n-2 k-1}+\cdots+z^{-2 n} ight\} d z \&=\frac{1}{2^{2 n} i} \cdot 2 \pi i\left(\begin{array}{c}2 n \n\end{array} ight)=\frac{1}{2^{2 n}}\left(\begin{array}{c}2 n \n\end{array} ight) 2 \pi \&=\frac{1}{2^{2 n}} \frac{(2 n) !}{n ! n !} 2 \pi=\frac{(2 n)(2 n-1)(2 n-2) \cdots(n)(n-1) \cdots 1}{2^{2 n} n ! n !} 2 \pi \&=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n} 2 \pi\end{aligned}[/latex]

Question 5.24: Prove that, for n = 1, 2, 3, ... , ∫^2π 0 cos^2n θ dθ = 1 . ...

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